\(\int \frac {\sin ^2(a+b \log (c x^n))}{x^3} \, dx\) [12]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 17, antiderivative size = 98 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {b^2 n^2}{4 \left (1+b^2 n^2\right ) x^2}-\frac {b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right ) x^2}-\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right ) x^2} \]

[Out]

-1/4*b^2*n^2/(b^2*n^2+1)/x^2-1/2*b*n*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))/(b^2*n^2+1)/x^2-1/2*sin(a+b*ln(c*x^
n))^2/(b^2*n^2+1)/x^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4575, 30} \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (b^2 n^2+1\right )}-\frac {b n \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{2 x^2 \left (b^2 n^2+1\right )}-\frac {b^2 n^2}{4 x^2 \left (b^2 n^2+1\right )} \]

[In]

Int[Sin[a + b*Log[c*x^n]]^2/x^3,x]

[Out]

-1/4*(b^2*n^2)/((1 + b^2*n^2)*x^2) - (b*n*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]])/(2*(1 + b^2*n^2)*x^2) -
 Sin[a + b*Log[c*x^n]]^2/(2*(1 + b^2*n^2)*x^2)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4575

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> Simp[(m + 1)*(e*x)^
(m + 1)*(Sin[d*(a + b*Log[c*x^n])]^p/(b^2*d^2*e*n^2*p^2 + e*(m + 1)^2)), x] + (Dist[b^2*d^2*n^2*p*((p - 1)/(b^
2*d^2*n^2*p^2 + (m + 1)^2)), Int[(e*x)^m*Sin[d*(a + b*Log[c*x^n])]^(p - 2), x], x] - Simp[b*d*n*p*(e*x)^(m + 1
)*Cos[d*(a + b*Log[c*x^n])]*(Sin[d*(a + b*Log[c*x^n])]^(p - 1)/(b^2*d^2*e*n^2*p^2 + e*(m + 1)^2)), x]) /; Free
Q[{a, b, c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 + (m + 1)^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right ) x^2}-\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right ) x^2}+\frac {\left (b^2 n^2\right ) \int \frac {1}{x^3} \, dx}{2 \left (1+b^2 n^2\right )} \\ & = -\frac {b^2 n^2}{4 \left (1+b^2 n^2\right ) x^2}-\frac {b n \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right ) x^2}-\frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+b^2 n^2\right ) x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {1+b^2 n^2-\cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )}{4 \left (1+b^2 n^2\right ) x^2} \]

[In]

Integrate[Sin[a + b*Log[c*x^n]]^2/x^3,x]

[Out]

-1/4*(1 + b^2*n^2 - Cos[2*(a + b*Log[c*x^n])] + b*n*Sin[2*(a + b*Log[c*x^n])])/((1 + b^2*n^2)*x^2)

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.60

method result size
parallelrisch \(\frac {-b^{2} n^{2}-b n \sin \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )+\cos \left (2 b \ln \left (c \,x^{n}\right )+2 a \right )-1}{4 x^{2} \left (b^{2} n^{2}+1\right )}\) \(59\)

[In]

int(sin(a+b*ln(c*x^n))^2/x^3,x,method=_RETURNVERBOSE)

[Out]

1/4*(-b^2*n^2-b*n*sin(2*b*ln(c*x^n)+2*a)+cos(2*b*ln(c*x^n)+2*a)-1)/x^2/(b^2*n^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {b^{2} n^{2} + 2 \, b n \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - 2 \, \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 2}{4 \, {\left (b^{2} n^{2} + 1\right )} x^{2}} \]

[In]

integrate(sin(a+b*log(c*x^n))^2/x^3,x, algorithm="fricas")

[Out]

-1/4*(b^2*n^2 + 2*b*n*cos(b*n*log(x) + b*log(c) + a)*sin(b*n*log(x) + b*log(c) + a) - 2*cos(b*n*log(x) + b*log
(c) + a)^2 + 2)/((b^2*n^2 + 1)*x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.44 (sec) , antiderivative size = 468, normalized size of antiderivative = 4.78 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\begin {cases} - \frac {3 i \sin {\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )} \cos {\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 x^{2}} - \frac {\cos ^{2}{\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{2 x^{2}} + \frac {\log {\left (c x^{n} \right )} \sin ^{2}{\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 n x^{2}} - \frac {i \log {\left (c x^{n} \right )} \sin {\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )} \cos {\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{2 n x^{2}} - \frac {\log {\left (c x^{n} \right )} \cos ^{2}{\left (a - \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 n x^{2}} & \text {for}\: b = - \frac {i}{n} \\- \frac {\sin ^{2}{\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{2 x^{2}} - \frac {i \sin {\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )} \cos {\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 x^{2}} + \frac {\log {\left (c x^{n} \right )} \sin ^{2}{\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 n x^{2}} + \frac {i \log {\left (c x^{n} \right )} \sin {\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )} \cos {\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{2 n x^{2}} - \frac {\log {\left (c x^{n} \right )} \cos ^{2}{\left (a + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}}{4 n x^{2}} & \text {for}\: b = \frac {i}{n} \\- \frac {b^{2} n^{2} \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} x^{2} + 4 x^{2}} - \frac {b^{2} n^{2} \cos ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} x^{2} + 4 x^{2}} - \frac {2 b n \sin {\left (a + b \log {\left (c x^{n} \right )} \right )} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} x^{2} + 4 x^{2}} - \frac {2 \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{4 b^{2} n^{2} x^{2} + 4 x^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(a+b*ln(c*x**n))**2/x**3,x)

[Out]

Piecewise((-3*I*sin(a - I*log(c*x**n)/n)*cos(a - I*log(c*x**n)/n)/(4*x**2) - cos(a - I*log(c*x**n)/n)**2/(2*x*
*2) + log(c*x**n)*sin(a - I*log(c*x**n)/n)**2/(4*n*x**2) - I*log(c*x**n)*sin(a - I*log(c*x**n)/n)*cos(a - I*lo
g(c*x**n)/n)/(2*n*x**2) - log(c*x**n)*cos(a - I*log(c*x**n)/n)**2/(4*n*x**2), Eq(b, -I/n)), (-sin(a + I*log(c*
x**n)/n)**2/(2*x**2) - I*sin(a + I*log(c*x**n)/n)*cos(a + I*log(c*x**n)/n)/(4*x**2) + log(c*x**n)*sin(a + I*lo
g(c*x**n)/n)**2/(4*n*x**2) + I*log(c*x**n)*sin(a + I*log(c*x**n)/n)*cos(a + I*log(c*x**n)/n)/(2*n*x**2) - log(
c*x**n)*cos(a + I*log(c*x**n)/n)**2/(4*n*x**2), Eq(b, I/n)), (-b**2*n**2*sin(a + b*log(c*x**n))**2/(4*b**2*n**
2*x**2 + 4*x**2) - b**2*n**2*cos(a + b*log(c*x**n))**2/(4*b**2*n**2*x**2 + 4*x**2) - 2*b*n*sin(a + b*log(c*x**
n))*cos(a + b*log(c*x**n))/(4*b**2*n**2*x**2 + 4*x**2) - 2*sin(a + b*log(c*x**n))**2/(4*b**2*n**2*x**2 + 4*x**
2), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (92) = 184\).

Time = 0.23 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.86 \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {2 \, {\left (b^{2} \cos \left (2 \, b \log \left (c\right )\right )^{2} + b^{2} \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} n^{2} + 2 \, \cos \left (2 \, b \log \left (c\right )\right )^{2} + {\left ({\left (b \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (c\right )\right ) - b \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + b \sin \left (2 \, b \log \left (c\right )\right )\right )} n - \cos \left (4 \, b \log \left (c\right )\right ) \cos \left (2 \, b \log \left (c\right )\right ) - \sin \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) - \cos \left (2 \, b \log \left (c\right )\right )\right )} \cos \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) + 2 \, \sin \left (2 \, b \log \left (c\right )\right )^{2} + {\left ({\left (b \cos \left (4 \, b \log \left (c\right )\right ) \cos \left (2 \, b \log \left (c\right )\right ) + b \sin \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + b \cos \left (2 \, b \log \left (c\right )\right )\right )} n + \cos \left (2 \, b \log \left (c\right )\right ) \sin \left (4 \, b \log \left (c\right )\right ) - \cos \left (4 \, b \log \left (c\right )\right ) \sin \left (2 \, b \log \left (c\right )\right ) + \sin \left (2 \, b \log \left (c\right )\right )\right )} \sin \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right )}{8 \, {\left ({\left (b^{2} \cos \left (2 \, b \log \left (c\right )\right )^{2} + b^{2} \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} n^{2} + \cos \left (2 \, b \log \left (c\right )\right )^{2} + \sin \left (2 \, b \log \left (c\right )\right )^{2}\right )} x^{2}} \]

[In]

integrate(sin(a+b*log(c*x^n))^2/x^3,x, algorithm="maxima")

[Out]

-1/8*(2*(b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2 + 2*cos(2*b*log(c))^2 + ((b*cos(2*b*log(c))*sin(4*
b*log(c)) - b*cos(4*b*log(c))*sin(2*b*log(c)) + b*sin(2*b*log(c)))*n - cos(4*b*log(c))*cos(2*b*log(c)) - sin(4
*b*log(c))*sin(2*b*log(c)) - cos(2*b*log(c)))*cos(2*b*log(x^n) + 2*a) + 2*sin(2*b*log(c))^2 + ((b*cos(4*b*log(
c))*cos(2*b*log(c)) + b*sin(4*b*log(c))*sin(2*b*log(c)) + b*cos(2*b*log(c)))*n + cos(2*b*log(c))*sin(4*b*log(c
)) - cos(4*b*log(c))*sin(2*b*log(c)) + sin(2*b*log(c)))*sin(2*b*log(x^n) + 2*a))/(((b^2*cos(2*b*log(c))^2 + b^
2*sin(2*b*log(c))^2)*n^2 + cos(2*b*log(c))^2 + sin(2*b*log(c))^2)*x^2)

Giac [F]

\[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {\sin \left (b \log \left (c x^{n}\right ) + a\right )^{2}}{x^{3}} \,d x } \]

[In]

integrate(sin(a+b*log(c*x^n))^2/x^3,x, algorithm="giac")

[Out]

integrate(sin(b*log(c*x^n) + a)^2/x^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^2\left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {{\sin \left (a+b\,\ln \left (c\,x^n\right )\right )}^2}{x^3} \,d x \]

[In]

int(sin(a + b*log(c*x^n))^2/x^3,x)

[Out]

int(sin(a + b*log(c*x^n))^2/x^3, x)